LMS (Least Mean Square) algorithm in the prior art adopts a single-filter structure as shown in FIG. 1. As shown in FIG. 2, its principle is that a signal received from one of the microphones is filtered, and the filtered signal is subtracted by a signal received from the other microphone to obtain a voice with noises reduced. The filter of the single-filter structure is merely updated in noise segments but remains unchanged in noisy voice segments.
If standard time domain LMS algorithm is used to compute convolutional non-additivity interference noises, the computation will be relatively complicated. In order to reduce the computational complexity, Ferrara proposed FBLMS (Fast Block LMS) algorithm, using a method of combining time and frequency domains, i.e., converting the original convolution operation in a time domain into a product operation in a frequency domain, which greatly reduces the computational complexity.
Hereinafter, the defects in LMS algorithm of the single-filter structure in the prior art will be described.
The defects in the single-filter structure will be expounded by analyzing the theoretical optimal solution of the filter in the single-filter structure. The analysis and calculation of the theoretical optimal solution of a filter is conducted in a frequency domain since the optimal solution of the filter can be clearly analyzed in a frequency domain.
FIG. 3 shows the analysis of the optimal solution of a filter frequency domain in a single-filter structure. In FIG. 3, S1 represents a signal source and S2 represents a noise source. Since FIR (Finite Impulse Response) filter can indicate more accurately the transfer function from a signal source to microphones, in the analysis, FIR filters are used to simulate the channel transfer function H11 between a signal source and a first microphone, the channel transfer function H12 between a noise source and the first microphone, the channel transfer function H21 between the signal source and a second microphone, and the channel transfer function H22 between the noise source and the second microphone, respectively. The signal received by the first microphone is X1, and the signal received by the second microphone is X2, W is a filter, and Y1 is a signal with noises reduced.
The following equations can be obtained:
                              X          ⁢                                          ⁢          1                =                              S            ⁢                                                  ⁢            1            ×            H            ⁢                                                  ⁢            11                    +                      S            ⁢                                                  ⁢            2            ×            H            ⁢                                                  ⁢            12                                              Equation        ⁢                                  ⁢        1                                          X          ⁢                                          ⁢          2                =                              S            ⁢                                                  ⁢            1            ×            H            ⁢                                                  ⁢            21                    +                      S            ⁢                                                  ⁢            2            ×            H            ⁢                                                  ⁢            22                                              Equation        ⁢                                  ⁢        2                                                                                    Y                ⁢                                                                  ⁢                1                            =                            ⁢                                                X                  ⁢                                                                          ⁢                  1                                -                                  X                  ⁢                                                                          ⁢                  2                  ×                  W                                                                                                        =                            ⁢                                                (                                                            S                      ⁢                                                                                          ⁢                      1                      ×                      H                      ⁢                                                                                          ⁢                      11                                        +                                          S                      ⁢                                                                                          ⁢                      2                      ×                      H                      ⁢                                                                                          ⁢                      12                                                        )                                -                                                                                                      ⁢                                                (                                                            S                      ⁢                                                                                          ⁢                      1                      ×                      H                      ⁢                                                                                          ⁢                      21                                        +                                          S                      ⁢                                                                                          ⁢                      2                      ×                      H                      ⁢                                                                                          ⁢                      22                                                        )                                ×                W                                                                                        =                            ⁢                                                S                  ⁢                                                                          ⁢                  1                  ×                                      (                                                                  H                        ⁢                                                                                                  ⁢                        11                                            -                                              H                        ⁢                                                                                                  ⁢                        21                        ×                        W                                                              )                                                  +                                  S                  ⁢                                                                          ⁢                  2                  ×                                                                                                                      ⁢                              (                                                      H                    ⁢                                                                                  ⁢                    12                                    -                                      H                    ⁢                                                                                  ⁢                    22                    ×                    W                                                  )                                                                        Equation        ⁢                                  ⁢        3            
Since noise source S2 will be completely eliminated when W is taken as the optimal solution, it can be inferred that the optimal solution of W is as shown in Equation 4:H12−H22×W=0W=H12/H22  Equation 4Y1=S1×(H11−H21×W)=S1×(H11−H21×H12/H22)  Equation 5
From Equation 5, it can be known that Y1 is a form of S1 that has been filtered in a certain mode and does not contain any component of S2.
From the above-obtained optimal solution as W=H12/H22, it can be seen that the optimal solution of W is not a FIR filter. Nevertheless, in practice, in order to ensure the stability and easy realization of a filter, a FIR filter is usually used, though it may introduce a great error because a non-FIR filter cannot be well approached by a FIR filter.
In a standard single-filter structure LMS algorithm, the optimal solution of a filter is a non-FIR filter. However, in practical application, the filter in this structure usually uses a FIR filter to approach this optimal solution, which may introduce a great error and cause poor noise elimination effect.